The half-life of carbon-14 is 5600 years. If charred logs from an old log cabin show only 71% of the carbon-14 expected in living matter, when did the cabin burn down?

When dealing with a half life question, it is best to use the half-life formula, which is expressed as:

`color(blue)(|bar(ul(color(white)(a/a)y=a(b)^(t/h)color(white)(a/a)|)))`

where:
`y=`final amount
`a=`inital amount
`b=`growth/decay
`t=`time elapsed
`h=`half-life

`1`. Start by expressing `71%` as a decimal and plugging the value into the half-life formula as well as your other known values.

• `a=1` is `100%`, expressed as a decimal
• `b=1/2` indicates half-life

`y=a(b)^(t/h)`

`0.71=1(1/2)^(t/5600)`

`2`. Since the bases are not the same on both sides of the equation, take the logarithm of both sides.

`log(0.71)=log((1/2)^(t/5600))`

`3`. Use the logarithmic property, `log_color(purple)b(color(red)m^color(blue)n)=color(blue)n*log_color(purple)b(color(red)m)`, to simplify the right side of the equation.

`log(0.71)=(t/5600)log(1/2)`

`4`. Isolate for `t`.

`t/5600=(log(0.71))/(log(1/2))`

`t==(5600log(0.71))/(log(1/2))`

`5`. Solve.

`color(green)(|bar(ul(color(white)(a/a)t~~2767.01color(white)(i)"years old"color(white)(a/a)|)))`